Question

The equilibrium constant (K) for the reaction
`Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be
[Given, `E_(cell)^(@)=0.46 V`]
A. `K_(C) = AL (15.6)`
B. `K_(C) = AL(2.5)`
C. `K_(C) = AL (1.5)`
D. `K_(C) = AL (12.2)`

Answer 1

Correct Answer - A
`E^(@) = (0.0592)/(n) "log" K_(C)`
`therefore log K_(C) = (E^(@) xx n)/(0.0592) = (0.46 xx 2 )/(0.0592) = 15.6`
`therefore K_(C) = AL (15.6)`